STK3405 Week 39 - uio.no · STK3405 – Week 39 A. B. Huseby & K. R. Dahl Department of Mathematics University of Oslo, Norway A. B. Huseby & K. R. Dahl (Univ. of Oslo) STK3405 –

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STK3405 – Week 39

A. B. Huseby & K. R. Dahl

Department of MathematicsUniversity of Oslo, Norway

A. B. Huseby & K. R. Dahl (Univ. of Oslo) STK3405 – Week 39 1 / 37

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Chapter 5

Structural and reliability importance for components inbinary monotone systems


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Importance measures

A measure of importance can be used to identify components thatshould be improved in order to increase the system reliability.

A measure of importance can be used to identify components thatmost likely have failed, given that the system has failed.


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Section 5.1

Structural importance of a component


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Criticality

Definition (Criticality)

Let (C, φ) be a binary monotone system, and let i ∈ C. We say thatcomponent i is critical for the system if:

φ(1i ,x) = 1 and φ(0i ,x) = 0.

If this is the case, we also say that (·i ,x) is a critical vector forcomponent i.

NOTE: Criticality is strongly related to the notion of relevance: Acomponent i in a binary monotone system (C, φ) is relevant if and onlyif there exists at least one critical vector for i .


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Criticality (cont.)

2 3 4

1

Figure: A binary monotone system (C, φ)

The structure function of the system (C, φ) is given by:

φ(x) = x1 q (x2 · x3 · x4)


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Criticality (cont.)

Component 1 is critical if (·1,x) is:

(·,0,0,0), (·,1,0,0), (·,0,1,0), (·,0,0,1),

(·,1,1,0), (·,1,0,1), (·,0,1,1).

Component 2 is critical if (·2,x) = (0, ·,1,1),

Component 3 is critical if (·3,x) = (0,1, ·,1),

Component 4 is critical if (·4,x) = (0,1,1, ·).


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Structural importance

Based on this Birnbaum suggested the following measure of structuralimportance of a component in a binary monotone system:

Definition (Structural importance)

Let (C, φ) be a binary monotone system of order n, and let i ∈ C. TheBirnbaum measure for the structural importance of component i, denoted J(i)

B ,is defined as:

J(i)B :=

12n−1

∑(·i ,x )

[φ(1i ,x)− φ(0i ,x)].

Note that the denominator, 2n−1 is the total number of states for the n − 1other components. Thus, J(i)

B can be interpreted as the fraction of all statesfor the n − 1 other components where component i is critical.


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2 3 4

1


For this system we have the following structural importance measures:

J(1)B =

724−1 =

78, J(2)

B = J(3)B = J(4)

B =1

24−1 =18.


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Let φ be a 2-out-of-3 system. To compute the structural importance ofcomponent 1, we note that the critical vectors for this component are(·,1,0) and (·,0,1). Hence, we have:

J(1)B =

223−1 =

12.

By similar arguments, we find that:

J(2)B = J(3)

B =12.

So in a 2-out-of-3 system, all of the components have the samestructural importance. This is intuitively obvious since the structurefunction is symmetrical with respect to the components.


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Section 5.2

Reliability importance of a component


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Reliability importance of a component

Definition (Reliability importance of a component)

Let (C, φ) be a binary monotone system, and let i ∈ C. Moreover, let Xbe the vector of component state variables.

The Birnbaum measure for the reliability importance of component i,denoted I(i)B is defined as:

I(i)B := P(Component i is critical for the system)

= P(φ(1i ,X )− φ(0i ,X ) = 1).


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Reliability importance of a component (cont.)

Since the difference φ(1i ,X )− φ(0i ,X ) is a binary variable, it followsthat:

I(i)B = E [φ(1i ,X )− φ(0i ,X )] = E [φ(1i ,X )]− E [φ(0i ,X )].

In particular, if the component state variables of the system areindependent, and P(Xi = 1) = pi for i ∈ C, we get that:

I(i)B = h(1i ,p)− h(0i ,p).


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Reliability importance of a component (cont.)

Theorem (Partial derivative formula)

Let (C, φ) be a binary monotone system where the component state variablesare independent, and P(Xi = 1) = pi for i ∈ C.

Then:I(i)B =

∂h(p)∂pi

, for all i ∈ C.

PROOF: By pivotal decomposition we have:

h(p) = pih(1i ,p) + (1− pi)h(0i ,p)

By differentiating this identity with respect to pi we get:

∂h(p)∂pi

= h(1i ,p)− h(0i ,p).

Hence, the result follows.A. B. Huseby & K. R. Dahl (Univ. of Oslo) STK3405 – Week 39 14 / 37

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Reliability importance inequalities

Theorem (Reliability importance inequalities)

For a binary monotone system, (C, φ), we always have

0 ≤ I(i)B ≤ 1.

Assume that the component state variables are independent, andP(Xj = 1) = pj , where 0 < pj < 1 for all j ∈ C.

If component i is relevant, we have:

0 < I(i)B .

Furthermore, if there exists at least one other relevant component, wealso have:

I(i)B < 1.


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Reliability importance inequalities (cont.)

PROOF: We note that the first inequality follows directly from thedefinition since the reliability importance is a probability.

We then assume that the component state variables are independent,and that P(Xj = 1) = pj , where 0 < pj < 1 for all j ∈ C.

If component i is relevant, we know that h is strictly increasing in pi .

That is, we must have:∂h(p)∂pi

> 0.

Combining this with the partial derivative formula, we get that 0 < I(i)B .


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Finally, we assume that there exists at least one other relevantcomponent k ∈ C.

To show that this implies that I(i)B < 1, we assume instead that I(i)B = 1,and show that this leads to a contradiction.

By this assumption, it follows that :

P(φ(1i ,X )− φ(0i ,X ) = 1) = 1

Since 0 < pj < 1, for all j ∈ C, it follows that P((·i ,X ) = (·i ,x)) > 0 forall (·i ,x).

Hence, we must have that:

φ(1i ,x) = 1 and φ(0i ,x) = 0 for all (·i ,x).


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At the same time, since component k is relevant, there exists a vector(·k ,y) such that:

φ(1k ,y) = 1 and φ(0k ,y) = 0.

If yi = 1, it follows that φ(1i ,0k ,y) = 0, contradicting that φ(1i ,x) = 1for all (·i ,x).

If yi = 0, it follows that φ(0i ,1k ,y) = 1, contradicting that φ(0i ,x) = 0for all (·i ,x).

Hence, we conclude that for both possible values of yi we end up withcontradictions.

Thus, the only possibility is that I(i)B < 1.


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Reliability importance and structural importance

Theorem (Reliability importance and structural importance)

Consider a binary monotone system (C, φ) where the component statevariables are independent, and where P(Xi = 1) = 1

2 for all i ∈ C.Then we have:

I(i)B = J(i)B

PROOF: If the component state variables are independent, andP(Xi = 1) = 1

2 for all i ∈ C, we have:

P((·i ,X ) = (·i ,x)) =∏j 6=i

P(Xj = xj) =∏j 6=i

(12) =

12n−1 .

From this the result follows.


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Reliability importance examples

In the following examples we consider binary monotone systems (C, φ)where C = {1, . . . ,n}.

We also assume that the component state variables are independent,and that:

P(Xi = 1) = pi , i ∈ C.

Without loss of generality we assume that the components are orderedso that:

p1 ≤ p2 ≤ . . . ≤ pn. (1)


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Reliability importance examples (cont.)

Let (C, φ) be a series system. Then for all i ∈ C we have:

I(i)B =∂∏n

j=1 pj

∂pi=

∏j 6=i

pj .

Hence, by the ordering (1), we get that:

I(1)B ≥ I(2)B ≥ · · · ≥ I(n)B .

Thus, in a series system the worst component, i.e., the one with thesmallest reliability, has the greatest reliability importance.


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Let (C, φ) be a parallel system. Then for all i ∈ C we have:

I(i)B =∂∐n

j=1 pj

∂pi=∂[1−

∏nj=1(1− pj)]

∂pi=

∏j 6=i

(1− pj).

Hence, from the ordering (1)

I(1)B ≤ I(2)B ≤ · · · ≤ I(n)B .

Thus, in a parallel system the best component, i.e., the one with thegreatest reliability, has the greatest reliability importance.


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Let (C, φ) be a 2-out-of-3 system. It is then easy to show that:

φ(X ) = X1X2 + X1X3 + X2X3 − 2X1X2X3.

Hence, we have:

h(p) = p1p2 + p1p3 + p2p3 − 2p1p2p3.

This implies that:

I(1)B =∂h(p)∂p1

= p2 + p3 − 2p2p3,

I(2)B =∂h(p)∂p2

= p1 + p3 − 2p1p3,

I(3)B =∂h(p)∂p3

= p1 + p2 − 2p1p2.


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We then consider the function f (p,q) = p + q − 2pq and note that:

I(1)B = f (p2,p3), I(2)B = f (p1,p3), I(3)B = f (p1,p2).

Moreover, the partial derivatives of f are respectively:

∂f∂p

= 1− 2q,∂f∂q

= 1− 2p.

If p,q ≤ 12 , f is non-decreasing in p and q. Thus, if p1 ≤ p2 ≤ p3 ≤ 1

2 , wehave:

f (p1,p2) ≤ f (p1,p3) ≤ f (p2,p3).

Hence, in this case we have:

I(3)B ≤ I(2)B ≤ I(1)B . (2)


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If p,q ≥ 12 , f is non-increasing in p and q. Thus, if 1

2 ≤ p1 ≤ p2 ≤ p3, wehave:

f (p2,p3) ≤ f (p1,p3) ≤ f (p1,p2).

Hence, in this case we have:

I(1)B ≤ I(2)B ≤ I(3)B . (3)


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If p1 = 12 − z, p2 = 1

2 and p3 = 12 + z, where z ∈ (0, 1

2 ), we get:

I(1)B = (12) + (

12+ z)− 2 · (1

2)(

12+ z) =

12,

I(2)B = (12− z) + (

12+ z)− 2 · (1

2− z)(

12+ z) =

12+ 2z2,

I(3)B = (12− z) + (

12)− 2 · (1

2− z)(

12) =

12,

Hence in this case we have:

I(1)B = I(3)B ≤ I(2)B . (4)

Note that this result holds also if z ∈ (− 12 ,0) in which case p1 > p2 > p3.


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2 3 4

1


The structure function of this system is:

φ(X ) = X1 q (X2 · X3 · X4) = X1 + X2 · X3 · X4 − X1 · X2 · X3 · X4

Thus, the reliability function is given by:

h(p) = p1 + p2 · p3 · p4 − p1 · p2 · p3 · p4


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Hence we have:

I(1)B = 1− p2 · p3 · p4

I(2)B = p3 · p4 − p1 · p3 · p4 = (1− p1) · p3 · p4

I(3)B = p2 · p4 − p1 · p2 · p4 = (1− p1) · p2 · p4

I(4)B = p2 · p3 − p1 · p2 · p3 = (1− p1) · p2 · p3

If p1 = p2 = p3 = p4 = p ∈ (0,1), we have:

I(1)B = 1− p3

I(i)B = p2 − p3 < I(1)B , i = 2,3,4.


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Assume instead that p1 = 0.1 and that p2 = p3 = p4 = 0.9. Then weget:

I(1)B = 1− p2 · p3 · p4 = 1− 0.93 = 0.271

I(2)B = p3 · p4 − p1 · p3 · p4 = (1− p1) · p3 · p4 = 0.93 = 0.729

I(3)B = p2 · p4 − p1 · p2 · p4 = (1− p1) · p2 · p4 = 0.93 = 0.729

I(4)B = p2 · p3 − p1 · p2 · p3 = (1− p1) · p2 · p3 = 0.93 = 0.729

Thus, in this case we have:

I(1)B < I(2)B = I(3)B = I(4)B .


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Section 5.3

The Barlow-Proschan measure of reliability importance


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The Barlow-Proschan measure of reliabilityimportance

Definition (Barlow-Proschan measure)

Consider a binary monotone system (C, φ), where the components arenever repaired.

Moreover, let Ti denote the lifetime of component i, i ∈ C, and let Sdenote the lifetime of the system.

The Barlow-Proschan measure of the reliability importance ofcomponent i ∈ C is defined as:

I(i)B−P := P(Component i fails at the same time as the system)

= P(Ti = S).


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Absolute continuity

A real-valued stochastic variable, T has an absolutely continuousdistribution if P(T ∈ A) = 0 for all measurable sets A ⊆ R such thatm1(A) = 0, where m1 denotes the Lebesgue measure in R.

If T1, . . . ,Tn are independent and absolutely continuously distributed,then T = (T1, . . . ,Tn) is absolutely continuously distributed in Rn.

That is, P(T ∈ A) = 0 for all (measurable) sets A ⊆ Rn such thatmn(A) = 0, where mn denotes the Lebesgue measure in Rn.

In particular, if A = {t : ti = tj}, where i 6= j , then mn(A) = 0.

Hence, P(Ti = Tj) = 0 when i 6= j .


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The Barlow-Proschan measure of reliabilityimportance (cont.)

Theorem (Probability of system failure)

Let (C, φ) be a non-trivial binary monotone system where thecomponents are never repaired and C = {1, . . . ,n}.Let Ti denote the lifetime of component i, i = 1, . . . ,n, and let S denotethe lifetime of the system.

Moreover, assume that T1, . . . ,Tn are independent, absolutelycontinuously distributed.

Then S is absolutely continuously distributed as well, and we have:

n∑i=1

I(i)B−P = 1.


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PROOF: Since we have assumed that the system is non-trivial, thelifetime of the system, S can be expressed as:

S = max1≤j≤p

mini∈Pj

Ti , (5)

where P1, . . . ,Pp are the minimal path sets of the system. This impliesthat:

P(n⋃

i=1

{Ti = S}) = 1. (6)

Let A ⊆ R be an arbitrary measurable set such that m1(A) = 0. Sincewe have assumed that T1, . . . ,Tn are absolutely coninuouslydistributed, we get that:

0 ≤ P(S ∈ A) ≤ P(n⋃

i=1

{Ti ∈ A}) ≤n∑

i=1

P(Ti ∈ A) = 0,

which implies that S is absolutely continuously distributed as well.A. B. Huseby & K. R. Dahl (Univ. of Oslo) STK3405 – Week 39 34 / 37

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Since T1, . . . ,Tn are absolutely continuously distributed, the probability ofhaving two or more components failing at the same time is zero.

This implies e.g., that P({Ti = S} ∩ {Tj = S}) = 0 for i 6= j . Thus, whencalculating the probability of the union of the events {Ti = S}, i = 1, . . . ,n, allintersections can be ignored as they have zero probability of occurring.

Hence, by (6) we get:

1 = P(n⋃

i=1

{Ti = S}) =n∑

i=1

P(Ti = S) =n∑

i=1

I(i)B−P ,

where the second equality follows by ignoring all intersections of events{Ti = S}, i = 1, . . . ,n.

The last equality follows by the definition of I(i)B−P , and hence, the proof iscomplete.


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Theorem (Integral formula for the Barlow-Proschan measure)

Let (C, φ) be a non-trivial binary monotone system where the componentsare never repaired, and where C = {1, . . . ,n}Let Ti denote the lifetime of component i, i = 1, . . . ,n.

Moreover, assume that T1, . . . ,Tn are independent, absolutely continuouslydistributed with densities f1, . . . , fn respectively.

Then, we have:

I(i)B−P =

∫ ∞0

I(i)B (t)fi(t)dt ,

where I(i)B (t) denotes the Birnbaum measure of the reliability importance ofcomponent 1 at time t, i.e., the probability that component i is critical at time t.


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PROOF: From the definitions of the Barlow-Proschan measure and theBirnbaum measure, it follows that:

I(i)B−P = P(Component i fails at the same time as the system)

=

∫ ∞0

P(Component i is critical at time t) · fi(t)dt

=

∫ ∞0

I(i)B (t)fi(t)dt .


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STK3405 Week 39 - uio.no · STK3405 – Week 39 A. B. Huseby & K. R. Dahl Department of Mathematics University of Oslo, Norway A. B. Huseby & K. R. Dahl (Univ. of Oslo) STK3405 –